栈的运用！ 表达式求值。

初级版本

http://acm.hdu.edu.cn/showproblem.php?pid=1237

拿杭电的这个初级版本练练手！

  1 #include<iostream>
  2 #include<algorithm>
  3 #include<stdio.h>
  4 #include<math.h>
  5 #include<string.h>
  6 #include<queue>
  7 #include<stack>
  8 using namespace std;
  9 #define maxn 17
 10
 11 stack <char> opS;//符号栈
 12 stack <double> numS;//数字栈
 13 char str[300];
 14 int pos[5] = {‘+‘,‘-‘, ‘*‘, ‘/‘, ‘#‘ };
 15 int cmps[5][5] = { {1,1,2,2,1},
 16                    {1,1,2,2,1},
 17                    {1,1,1,1,1},
 18                    {1,1,1,1,1},
 19                    {2,2,2,2,3} };
 20
 21 char* getNum(char *p,double &num)
 22 {
 23     num = 0;
 24     while(*p != ‘ ‘ && *p)
 25     {
 26         num = num*10 + *p - ‘0‘;
 27         p ++;
 28     }
 29
 30     if(*p == ‘ ‘)
 31         p ++;
 32
 33     return p;
 34 }
 35 int getOp(char ch)
 36 {
 37     for(int i=0; i<5; i++)
 38     {
 39         if(ch == pos[i])
 40             return i;
 41     }
 42 }
 43
 44
 45 int cmp(char op2,char op)
 46 {
 47     int a, b;
 48     a = getOp(op2);
 49     b = getOp(op);
 50
 51     return cmps[a][b];
 52 }
 53
 54 double cal(double a, double b,char op)
 55 {
 56     switch(op)
 57     {
 58         case ‘+‘:return a+b;
 59         case ‘-‘:return b-a;
 60         case ‘*‘:return a*b;
 61         case ‘/‘:return b/a;
 62     }
 63
 64 }
 65
 66 double solve()
 67 {
 68     double num1, num2;
 69     char op, *p = str, op2;
 70
 71     opS.push(‘#‘);
 72
 73     while(*p)
 74     {
 75         p = getNum(p,num1);
 76
 77         if(*p)
 78         {
 79             op = *p;
 80             p += 2;
 81         }
 82         else
 83             op = ‘#‘;
 84
 85         numS.push(num1);
 86
 87         while(1)
 88         {
 89             op2 = opS.top();
 90
 91             if(cmp(op2,op) == 2)///如果栈内运算符的优先级 比栈外的优先级要低 操作符都入栈
 92             {
 93                 opS.push(op);
 94                 break;
 95             }
 96             else if(cmp(op2, op) == 3)
 97             {
 98                 return numS.top();
 99             }
100             else
101             {
102                 num1 = numS.top();
103                 numS.pop();
104                 num2 = numS.top();
105                 numS.pop();
106                 opS.pop();
107
108                 num1 = cal(num1, num2, op2);
109                 numS.push(num1);
110             }
111         }
112
113
114     }
115
116 }
117
118 int main()
119 {
120     while(gets(str), strcmp(str, "0") )
121     {
122         while(opS.size() )
123             opS.pop();
124         while(numS.size() )
125             numS.pop();
126
127         double ans = solve();
128
129         printf("%.2lf\n", ans);
130     }
131     return 0;
132 }